J. Miller Posted February 24, 2008 Share Posted February 24, 2008 Hopefully this is an easy one, but it's got me stumped. I can find the begining of my Linear PIO with getSymLoc, but I am trying to find the other end of my Linear PIO (its ending point). I am looking for its ending x,y location. Any suggestions. Thanks in Advance Jeff Miller Quote Link to comment
Pat Stanford Posted February 25, 2008 Share Posted February 25, 2008 GetSymRot will get you the angle. The parameter record had the length. A little Trigonometry will get you the X and Y of the end point. Pat Quote Link to comment
Miguel Barrera Posted February 25, 2008 Share Posted February 25, 2008 Alternatively, you can use a vector to get the end point with the function Ang2Vec(angleR :REAL;Length :REAL):VECTOR; endVec:= Ang2Vec(angleR,Length); endPt.x:= begPt.x + endVec.x; endPt.y:= begPt.x + endVec.y; Quote Link to comment
Miguel Barrera Posted February 25, 2008 Share Posted February 25, 2008 last operation should be: endPt.y:= begPt.y + endVec.y; Quote Link to comment
J. Miller Posted February 25, 2008 Author Share Posted February 25, 2008 Thanks guys, I thought about that method driving into work today. At least I know I'm heading in the right direction now Jeff Miller Quote Link to comment
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